# Limit of the absolute value of a general function

Given the function $f: X \rightarrow \mathbb{R}$ with $\lim_{x\rightarrow a}f(x) = L$. Prove that $\lim_{x\rightarrow a}|f(x)| = |L|$.

$\forall \epsilon > 0 \exists\delta>0 s.t. 0<|x-a|< \delta \Rightarrow ||f(x)| - |L||\leq |f(x) - L| < \epsilon$.

I'm not really sure regarding this step $$||f(x)| - |L||\leq |f(x) - L| < \epsilon$$

Is it correct?

• When you write $||f(x) - |L||$ do you mean $||f(x)| - |L||$ or do you genuinely think it's the former? – Zain Patel May 10 '17 at 3:08
• @ZainPatel, it was a typo. I meant $||f(x)| -|L||$. – KirkLand May 10 '17 at 22:18

you have it nearly correct. the inequality you need is (with $a=f(x)$ and $b=L$) $$||a|-|b|| \le |a-b|$$ which holds in fact for any complex numbers $a,b$. it states that the minimum distance between two concentric circles is no greater that the distance separating any two points, one on each circle.